I have $$yf'_x-xf'_y=0$$ and should solve it using the variable substitutions $$u=x^2+y^2, v=x$$
I have managed to get $$f'_x=2xf'_u+f'_v$$ and $$f'_y=2yf'_u$$ through the chain rule.
This in the original equation gives me $$yf'_v=0$$
The solution provided is as below but I don't understand it.
So $f=h(u)=h(x^2+y^2)$ where $h$ $C'$ function of a variable. (Handwritten, might have typos)
I need help with understanding what the suggested solution means.
according to your variables $u=x^2+y^2$ and $v=x$ you have a simplified equation: $$y\frac{\partial f}{\partial v}=0$$ Since this equation must always hold (for any arbitrary value of $y$) the derivative of $f$ wrt. $v$ must be zero, and therefore $f$ must be a function of $u$ only: $$f=h(u) = h(x^2+y^2)$$
Another approach is:
You can put your equation for $f(x,y)$: $$ y\frac{\partial f(x,y)}{\partial x}-x\frac{\partial f(x,y)}{\partial y}=0\tag{*}$$ as the total derivative of $f$ wrt. some parameter $t$ supposing: $$x=x(t) \qquad y=y(t)$$ equated to $0$. This is: $$\frac{df(x(t),y(t))}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}=0\tag{**}$$
You can immediately see that equations $(*)$ and $(**)$ have a lot in common, and they are identical iff: $$\frac{dx}{dt} = y \qquad \frac{dy}{dt}=-x$$ The latter equations can be multiplied by $x$ and $y$ respectively and adding them toguether gives: $$ x\frac{dx}{dt}+y\frac{dy}{dt}=\frac{d}{dt}\left(x^2+y^2\right) = 0$$
Therefore the taryectories described by curves on $xoy$ plane such as $x^2+y^2$ have a constant $f$ throughout the motion. Therefore, by equation $(**)$ the function $f$ must be a function of $x^2+y^2$, namely $f=h(x^2+y^2)$ since
$$\frac{df(x,y)}{dt} = h'(x,y)\frac{d}{dt}\left(x^2+y^2\right)=0$$